使用OrderedDict按项目长度订购python字典

Question

Im trying to use OrderedDict to make a dictionary where the item is a list of entries, and I want the ordering to be based on the length of the item list.

say for:

d = {2:[1], 1:[1,2,3], 4:[1,2], 5:[1,2,3,4]}

I tried:

od = collections.OrderedDict(sorted(len(d.items())))

but I got a TypeError: 'int' object is not iterable

I found a workaround by iterating something like this:

for k in sorted(d, key=lambda k: len(d[k]), reverse=True): 

when needed, but I'd rather the dict be sorted beforehand.

Answer 1

You could sort d.items() according to the length of the values :

In [268]: sorted(d.items(), key=lambda x: len(x[1]), reverse=True)
Out[268]: [(5, [1, 2, 3, 4]), (1, [1, 2, 3]), (4, [1, 2]), (2, [1])]

and pass that to collections.OrderedDict :

In [267]: collections.OrderedDict(sorted(d.items(), key=lambda x: len(x[1]), reverse=True))
Out[267]: OrderedDict([(5, [1, 2, 3, 4]), (1, [1, 2, 3]), (4, [1, 2]), (2, [1])])